Answer:
a) f(4) = 2(4) - 5 = 3
Step-by-step explanation:
First, we look at the function, and we see different domain restrictions.
We are given the value of 4. The first equation has a domain restriction which includes 4 as it has the line below the greater than sign.
So, the first equation must work.
So our answer is A.
If A is QIV, then 3π/2 ≤A≤2π;
we have to find out in what quadrant is A/2
(3π/2)/2≤A/2≤(2π)/2 ⇒ 3π/4≤A/2≤π
We can see that A/2 will be in QIII; therefore the sec (A/2) will be negative (-).
1) we have to calculate cos (A/2)
Cos (A/2)=⁺₋√[(1+cos A/2)/2]
We choose this formula: Cos (A/2)= -√[(1+cos A/2)/2], because sec A/2 is in quadrant Q III, and the secant (sec A/2=1/cos A/2) in this quadrant is negative.
Cos (A/2)=-√[(1+cos A)/2]=-√[(1+(1/2)]/2=-√(3/4)=-(√3)/2.
2) we compute the sec (A/2)
Data:
cos (A/2)=-(√3)/2
sec (A/2)=1/cos (A/2)
sec (A/2)=1/(-(√3)/2)=-2/√3=-(2√3)/3
Answer: sec (A/2)=-(2√3)/3
Answer:
8
Step-by-step explanation:
can i have brainlst
so at 8AM the temperature was less than -7C, so recall that on the negative side, the farther from 0, the smaller, so that number could be say -8C, or -9C or -10C or even say -20C, all those are lesser than -7C, and therefore colder as well.
let's say it was -20C, by noon it went up by 7C, so it became -20 + 7 = -13, well, -13C is still less than -7C, so that'd be true for the B) statement.
let's say it was -10C, by noon it went up by 7C, so it became -10 + 7 = -3, well, -3 is more than -7 and less than 0, so that'd be true for the D) statement.
Answer:
Yes. It is!
Step-by-step explanation:
JK is not a full line, it would be a ray.
